Question

For separation $r$ between gold nucleus and an $\alpha$-particle, the magnitude of repulsive force is given by

• A. $F=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{2 e^{2}}{r^{2}}$
• B. $F=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{e^{2}}{r^{2}}$
• C. $F=\frac{158 \cdot e^{2}}{4 \pi \varepsilon_{0} r^{2}}$
• D. $F=\frac{56 e^{2}}{4 \pi \varepsilon_{0} r^{2}}$

Answer 1

Answer: (C)

For separation $r$, force between $\alpha$-particle (charge $2 e$ ) and gold nucleus (charge $Z e$ ) is
$F=\frac{1}{4 \pi \varepsilon_{0}} \times \frac{2 e(Z e)}{r^{2}}$
For gold,
$Z=79$
$\therefore F=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{158 e^{2}}{r^{2}}$